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Antenna

Chapter 4

2

Chapter OutlinesChapter 4 Antenna

Fundamental Parameters of Antennas Electrically Short Antennas Dipole Antennas Monopole Antennas Antenna Arrays Helical Antennas Yagi Uda Array Antennas Antennas for Wireless Communications

3

IntroductionWires passing an alternating current emit or radiate EM energy. The shape and size of the current carrying structure determine how much energy is radiated, direction of propagation and how well the radiation is captured.

Henceforth, the structure to efficiently radiate in a preferred direction is called transmitting antenna, while the other side which is to capture radiation from preferable direction is called receiving antenna.In most cases, the efficiency and directional nature for an antenna is the same whether its receiving or transmitting.

4

Common types of antennas:

Introduction (Cont’d..)

Fundamentals of Electromagnetics With Engineering Applications by Stuart M. WentworthCopyright © 2005 by John Wiley & Sons. All rights reserved.

Figure 8-1 (p. 389)Common single-element antennas.

5

Introduction (Cont’d..)


Generic antenna network. The antenna acts as a transducer between guided waves on the T-line and waves propagating in space.

4.1 Fundamental Parameters of AntennasTo describe the performance of an antenna, definitions of various parameters are discussed. The radiated power, beam pattern, directivity, and efficiency are all important parameters in characterizing antenna.RADIATED POWER

Suppose transmitting antenna located at the origin of spherical coordinate. From this coordinate system, there are three components of radiated field, in r, θ and φ. But, the intensities of these components vary with radial distance as 1/r, 1/r2 and 1/r3.

Fundamental Parameters of Antennas (Cont’d..)


Figure 8-3 (p. 392) The spherical radiation pattern for an isotropic antenna.

For almost all practical applications, a receiving antenna located far enough away from the transmitter (as a point source of radiation) far field region.A distance r from the origin is generally accepted as being in the far field region if :

22Lr

L is the length of the largest dimension on the antenna element, and assumed L>λ. For smaller L, r should at least as large as λ

Fundamental Parameters of Antennas (Cont’d..)In the far field, the radiated waves resemble plane waves propagating in ar direction where time harmonic fields related by :

12010

00 EaH HaE SrSSrS

The time averaged power density vector of the wave is by Poynting Theorem,

*Re21,, SSr HEP

Where in the far field, rrPr a P ,,,,


So, the total power radiated by the antenna, Prad is found by integrating it over a close spherical surface :

ddrrPdSrPrad sin,,.,,P 2

10

Example 1

In free space, suppose a wave propagating radially away from an antenna at the origin has:

Where the driving current phasor jS eII 0

Find:• ES

• P(r,φ,θ) • Prad and Rrad

2cos ,ss

Ir

H a

Solution to Example 1

To find ES, from time harmonic fields relation:

So,

2

2

cos ,

( ) cos

ss P s o r

o ss

Ir

Iar

E a H a a

E a

* 2 2

24

1 1Re Re cos cos2 2

1(b) cos2

j jo o o

s s

oo r

I e I er r

Ir

r

P E H a a

P , , a

Then, to find P(r,φ,θ)

Solution to Example 1 (Cont’d..)Then,

2

2

2

2 1205 961

2( ) 950

o

rad

o

rad

IR

I

c R

Solving:

200

2

00

4200

22

42

00

20

52

sincos21

sincos21

21,,

I

ddI

ddrr

I

RIdrP

rr

radrad

aa

SP

Fundamental Parameters of Antennas (Cont’d..)RADIATION PATTERNSFrom Balanis book:

Fundamental Parameters of Antennas (Cont’d..)Radiation patterns usually indicate either electric field, E intensity or power intensity. Magnetic field intensity, H has the same radiation pattern as E related by η0. The polarization or orientation of the E field vector is an important consideration in an E field plot. A transmit receive antenna pair must share same polarization for the most efficient communication.


Coordinate System

Fundamental Parameters of Antennas (Cont’d..)Since the actual field intensity is not only depends on radial distance, but also on how much power delivered to antenna, we use and plot normalized function divide the field or power component with its maximum value.

E.g. the normalized power function or normalized radiation intensity :

max

,,,PrPPn


1, isonP So,

In contrast with isotropic antenna, a directional antenna radiates and receives preferentially in some direction.

If the antenna radiates EM waves equally in all directions, it is termed as isotropic antenna, where the normalized power function is equal to 1.



Figure 8-4a (p. 392)General far-field antenna radiation pattern: (a) polar plot; (b) rectangular plot.


Figure 8-4a (p. 392)General far-field antenna radiation pattern: (a) polar plot; (b) rectangular plot.

The normalized radiation patterns for a generic antenna, called polar plot. A 3D plot of radiation pattern can be difficult to generate and work with, so take slices of the pattern and generate 2D plots (rectangular plots) for all θ at φ=π/2 and φ=3π/2

Polar plot

Rectangular plot (in dB)


The polar plot also can be in terms of dB. Where normalized E field pattern,

max

,,,ErEEn

This will be identical to the power pattern in decibels if:

,log20, nn EdBE

whereas

,log10, nn PdBP


The are some zeros and nulls in radiation pattern, indicating no radiations. These lobes shows the direction of radiation, where main or major lobe lies in the direction of maximum radiation. The other lobes divert power away from the main beam, so that good antenna design will seek to minimize the side and back lobes.Beam’s directional nature is beamwidth, or half power beamwidth or 3 dB beamwidth. It will shows the angular width of the beam measured at the half power or -3 dB points.

Fundamental Parameters of Antennas (Cont’d..)For linearly polarized antenna, performance is often described in terms of its principal E and H plane patterns.E plane : the plane containing the E field vector & the direction of max radiationH plane : the plane containing the H field vector & the direction of max radiation

For next figure, • the x-z plane (elevation plane, φ=0) is the principal E-

plane• the x-y plane (azimuthal plane; θ=π/2) is the principal H-

plane.


For this figure, • Infinite number of

principal E-planes (elevation plane, φ=φc).• One principal H-plane

(azimuthal plane; θ=π/2).

Fundamental Parameters of Antennas (Cont’d..)DIRECTIVITY

A measure of how well an antenna radiate most of the power fed into the main lobe. Before defining directivity, describe first the antenna’s pattern solid angle or beam solid angle.


Figure 8-5 (p. 393)(a) An arc with length equal to a circle’s radius defines a radian. (b) An area equal to the square of a sphere’s radius defines a steradian.

An arc with length equal to a circle’s radius defines a radian.


An area equal to the square of a sphere’s radius defines a steradian (sr).A differential solid angle dΩ in sr is:

ddd sin

For sphere, the solid angle is found by integrating dΩ :

sr

4sin2

0 0

dd


An antenna’s solid angle Ωp:

dPnp ,

To find the normalized power’s average value taken over the entire spherical solid angle :

4

,, pn

avend

dPP

Fundamental Parameters of Antennas (Cont’d..)The directive gain D(θ,φ) of an antenna is the ratio of the normalized power in particular direction to the average normalized power :

aven

nP

PD

,

,,

The directivity Dmax is the maximum directive gain,

1,

,,

, maxmax

maxmax

naven

n PwherePP

DD

So,

pD

4

max

Fundamental Parameters of Antennas (Cont’d..)Directivity in decibels as:

maxmax log10 DdBD

Useful relation:

,,, max nPDD

Total radiated power as:

dPPrPrad ,max2

Or :prad PrP max

2


For (a), power gets radiated to the side and back lobes, so the pattern solid angle is large and the directivity is small. For (b), almost all the power gets radiated to the main beam, so pattern solid angle is small and directivity is high.

31

Example 2

2 3, sin for 0 ,

0 otherwise.

sin nP

For this normalized radiation intensity,

Find the beamwidth, pattern solid angle and the directivity.

32


The beam is pointing in the +y direction. WHY?!?!

Due to the beam having function in θ and φ, so that:

1 .2

BW BW BW

From previous equation, so that a 3dB beamwidth is at half of total power = 0.5.

1, max nP

33

Solution to Example 2 (Cont’d..)

To find BWθ, we fix φ = π/2 to get:

and then set sin2θ equal to ½. Then,

1sin, 2 nP

1 1sin 45 , so 180 45 45 90 .2

BW

To find BWφ, we fix θ = π/2 to get:

3sin1, nP

and then set sin3θ equal to ½. Then,


11 3sin 1 2 52.5 , so 180 52.5 52.5 75 .BW

So, 1 90 75 82.5 .

2BW

The pattern solid angle is:

2 3

3 3

0 0

sin sin sin ,

sin sin , (note limits on )

P n

P

P d d d

d d

35


Where each integral is solved as follows:

3 2 2

0 0 0 0

sin 1 cos sin sin cos sin .y xdx x xdx xdx x xdx

Please continue on your own!!

3

434sinsin

0 0

33

ddp

srp 78.1Finally,

So,

max4 4 7.1

1.78P

D

and the directivity,

Fundamental Parameters of Antennas (Cont’d..)IMPEDANCE & EFFICIENCY

(a) A T-line terminated in a dipole antenna can be modeled with an antenna impedance, Zant (b) consisting of resistive and reactive components (c).


The antenna resistance consists of radiation resistance Rrad and a dissipative resistance Rdiss that arises from ohmic losses in the metal conductor. For antenna driven by phasor current,

radrad RIP 202

1 and also dissdiss RIP 2

021

For maximum radiated power, Rrad need to be as large as possible but without being too large, for easily match with the feedline.

Fundamental Parameters of Antennas (Cont’d..)So then the antenna efficiency, e

dissrad

rad

dissrad

radRR

RPP

Pe

The power gain G(θ,φ) is likely its directive gain plus efficiency, where:

,, eDG

And the max power gain is when the directivity is max. It’s been always expressed in dBi , indicating dB with respect to an isotropic antenna.

4.2 Electrically Short Antennas

If the current distribution of a radiating element is known, it’s possible to calculate the radiated fields by a direct integration but, the integrals can be very complex.For time harmonic fields, integration is performed to find a phasor called retarded vector magnetic potential, which then followed by simple differentiation to find the H field. We will begin with a derivation of the retarded vector magnetic potential, then find the radiated fields for Hertzian dipole (infinitesimally short element with uniform current along its length). From that, we can find the fields from longer structures via integration e.g. small loop antenna.

Electrically Short Antennas (Cont’d..)Vector Magnetic Potential

In working with electric fields, and analogous term for H field is the vector magnetic potential A, often used for antenna calculations.

VE

Where in the point form of Gauss Law for magnetic fields,

0 BWith vector identity that states the div of the curl of any vector A is zero. So,

AB Then we now seek a relation between vector A and a current source.

Electrically Short Antennas (Cont’d..)From Biot Savart law,

ddo

dod dv

R 20

0 4a

JB

Remember?!?

The vector magnetic potential at the observation point (o) results from a current density distributed about the volume vd.

Electrically Short Antennas (Cont’d..)

Due to long derivation and by using vector identity, we could get:

ddo

d dvR 2

000 4

JB

Then,

ddo

d dvR 2

00 4

JA

More properly,

d

do

dddd dvR

zyxzyx

20

0000,,

4,,

JA

Where the vector

potential at point o is a function of the position of current element

Refer to text book!


We have yet to consider time dependence, so:

d

do

pdodddd dvR

uRtzyxtzyx

/,,,

4,,, 0

0000J

A

Where:

tjdspdodddd euRtzyx JJ /,,,

Where Jds is the retarded phasor quantity:

dojkRddddds ezyx ,,JJ k = β =

2π/λ


Using this current density, the phasor form is:

ddo

dsS dv

RJ

A4

00

This is what we called as the time harmonic equation for the retarded vector magnetic potential.

In phasor notation, the vector magnetic potential is related to the magnetic flux density by :

SS 00 AB


From there, we could find H in free space :

000 / SS BH

Because the radiation is propagating radially away from the source, it is then a simple matter to find E, where:

SrS 000 HaE

Finally, the time averaged power radiated is:

SSr 00Re21,, *HEP

Electrically Short Antennas (Cont’d..)Hertzian Dipole

Suppose that a short line of current, tIti cos)( 0

It’s placed along the z axis as shown.


Here, the phasor current is:To maintain constant current over its entire length, imagine a pair of plates at the ends of the line that can store charge.The stored charge at the ends resembles an electric dipole, and the short line of oscillating current is then ‘Hertzian Dipole’.

js eII 0

For Is in the +az direction through a cross sectional S, the current density at the source seen by the observation point:

zjkRs

ds eSI

aJ


Geometrical arrangement of an infinitesimal dipole and its associated electric field components on a spherical surface.

Electrically Short Antennas (Cont’d..)A differential volume of this element,

zjkR

sdds dzeIdv aJ

Sdzdvd So,

With assumption that the Hertzian dipole is very short, integrate to get A0S :

2/

2/

00 4

z

jkRs

S dzR

eIaA

Therefore,

z

jkRs

S ReI

aA

4

00

Electrically Short Antennas (Cont’d..)The unit vector az can be converted to its equivalent spherical coordinates, so that:

aaA sincos

40

0

r

jkRs

S ReI

It is now a relatively straightforward matter to find B0S and then,

aH sin1

40

Rjk

ReI jkR

sS

Regroup to get:

aH sin1

4 2

2

0

kRkRjekI jkR

sS

Electrically Short Antennas (Cont’d..)The second terms drops off with increasing radius much faster than the first term, where for far field condition:

211

kRkR

aH sin

40 RkeI

jjkR

sS

Therefore,

Meanwhile, the far field value of E field :

aE sin

400 RkeI

jjkR

sS

Electrically Short Antennas (Cont’d..)The time averaged power density at observation point is:

rR

kIr aP

2

22

22200 sin

32,

The term in brackets is the max power density. The sin2θ term is the normalized radiation intensity Pn(θ) plotted as :

Electrically Short Antennas (Cont’d..)The normalized radiation intensity can be used to find the pattern solid angle,

38sinsinsin 22 dddp

The directivity is then: 5.14max

pD

20

222

220

202

max2 40

32I

R

IkrPrP pprad

The total power radiated by a Hertzian dipole:


Also Rrad as:2

280

radR

For Hertzian dipole, where l<<λ, Rrad will be very small and the antenna will not efficiently radiate power. Larger dipole antennas, have much higher Rrad and thus more efficient.

55

Example 3

Suppose a Hertzian dipole antenna is 1 cm long and is excited by a 10 mA amplitude current source at 100 MHz. What is the maximum power density radiated by this antenna at a 1 km distance? What is the antenna’s radiation resistance?


8

6

3 10, 3 .100 10 1

c x m sc f mf x s

2 2 22 2 2

max 2 2 2 2 2 2

2 0.010 0.010120 0.05232 32 3 1000o oI pWP

r m

l

2 22 2 0.0180 80 8.8

3radR m l

We should calculate the wavelength,

The max. power radiated is:

The antenna radiation resistance :

Electrically Short Antennas (Cont’d..)Small Loop Antenna

A small loop of current located in the xy plane centered at the origin small loop antenna or magnetic dipole.


aJ adeIdv jkRsdds

Substitute,

Giving us,

aA dR

eaI jkRs

S 40

0

Assume that a<<λ and A0s is in far field. It’s discovered that,

aA sin1

4 20

0jkRs

S ejkRR

SI

Where, and thus,2aS

aH jkRsS e

RSkI

sin4 0

00

aE jkRsS e

RSkI sin

40

0

Electrically Short Antennas (Cont’d..)The power density vector :

rR

kSIr aP

2

220

2220

20

2sin

32,

Where,

220

2220

20

2

max32 R

kSIP

Since the normalized power function is the same as Hertzian dipole, Ωp=8π/3, Dmax = 1.5


2

2

20

30

34

SIPrad

Calculation of Prad and Rrad yields

2

24320

SRrad

The fields for the small loop antenna is similar to Hertzian dipole. It’s the dual of Hertzian (electric) dipole magnetic dipole.

The equations also valid for multi turn loop, as long as the loop small compared to wavelength. For N circular loop, S=Nπa2 and for square coil N loops, with each side length b, S=Nb2

Try this!!

4.3 Dipole Antennas

Dipole Antennas

A drawback to Hertzian dipole as a practical antenna is its small radiation resistance. A longer will have higher radiation resistance, becomes more efficient. It as an L long conductor conveniently placed along the z axis with current distribution i(z,t).

Assume sinusoidal current distribution on each arm, where the antenna is center-fed and the current vanishes at the end points. The distribution:

tzItzi s cos)(,

Where,

02/

2sin

2/02

sin)(

0

0

zLzLkeI

LzzLkeIzI

j

j

s

Division of the L long dipole into a series of infinitesimal Hertzian dipoles !!

Dipole Antennas (Cont’d..)

For simplicity, assume phase term =0, and make use of current distribution term with magnetic field equation for a Hertzian dipole to get :

2/

0

0

2/

00 'sin

2sin'sin

2sin

4

L jkR

L

jkR

S dzzL

kR

edzz

Lk

RekI

j aH


For far field, the vectors r and R appear to be parallel, so that θ’=θ and R=r, where:

coscos jkzjkrzrjkjkR eeee

After pulling the components that don’t change with z, use table of integrals and with application to Euler’s identity,

aaH

sin2

coscos2

cos

20

00

kLkL

reI

jHjkr

SS


The vector E0S is then easily found from :

a HaE SSrS H00000

Where, the pattern function is given by:

2

sin2

coscos2

cos

kLkL

F

The time averaged power radiated is:

rFr

Ir a P

2

2015

,


Therefore, the normalized power function is:

max

FFPn

And the max time averaged power density is then :

max2

20

max15

Fr

IP

It’s not generally equivalent to the normalized power function since F(θ) can be greater than one.


3D and 2D amplitude patterns for a thin dipole of l=1.25λ


Half Wave Dipole Antennas

Because of its convenient radiation resistance, and because it’s a smallest resonant dipole antenna, the half wavelength dipole antenna merits special attention.With kL/2 = π/2,

rr

Ir a P

2

2

2

20

sin

cos2

cos15

,

With the F(θ) is 1, the maximum power density is: 2

2015

,r

Ir

P


Therefore, the normalized power density is:

2

2

cos

cos2

cos

nP

The current distribution and normalized radiation pattern for a half wave dipole antenna.


For half wave dipole…658.7 p and with L = λ/2 Try

this!!We can then find the directivity as :

640.14max

pD

Which is slightly higher than the directivity of Hertzian dipole, the radiation resistance is given by:

pradrad PrRIP max22

021

Leads to: 2.7330

pradR


This radiation resistance much higher than of Hertzian dipole, where it radiates more efficiently easier to construct an impedance matching network for this antenna impedance.This antenna impedance also contains a reactive components, Xant, where for a λ/2 dipole antenna it is equal to 42.5Ω . Therefore, total impedance by neglecting Rdiss. 5.422.73 jZant

For impedance matching, need to make reactance zero (in resonant condition). So, it can be achieved by making the antenna slightly shorter (reduced in length until reactance vanishes).


74

Example 4

Find the efficiency and maximum power gain of a λ/2 dipole antenna constructed with AWG#20 (0.406 mm radius) copper wire operating at 1.0 GHz.

Compare your result with a 3 mm length dipole antenna (Hertzian Dipole) if the center of this antenna is driven with a 1.0 GHz sinusoidal current.


We first find the skin depth of copper at 1.0 GHz,

mfcu

6

7790

1009.2108.5104101

11

This is much smaller than the wire radius, so the wire area over which current is conducted by:

291033.52 maS cu

At 1 GHz, the wavelength is 0.3m and the λ/2 is 0.15m long. The ohmic resistance is then:

485.01S

Rdiss

Solution to Example 4 (Cont’d..)Since the radiation resistance for half wave dipole is 73.2Ω, we have :

99.0485.02.73

2.73

e

A gain of:

63.1640.199.0maxmax eDG

Meanwhile for Hertzian Dipole, the ohmic resistance of the small dipole is :

mS

Rdiss 7.91

Solution to Example 4 (Cont’d..)To find the radiation resistance, with value of wavelength is 0.3m, thus :

m

mm

Rrad 793.0

10380

32

Therefore, its efficiency and gain:

89.07.979

79

mm

me 34.15.189.0maxmax eDG

Thus, the half wave dipole is clearly more efficient with a higher gain than the short dipole.

Consider the construction of half wave dipole for an AM radio station broadcasting at 1 MHz. At this f, the wavelength is 300m long and the half wave dipole antenna must be 150m tall.We can cut this in half, by employing image theory to build a quarter wave monopole antenna that is only 75m tall!!

4.4 Monopole Antennas

Monopole Antennas (Cont’d..)

Consider pair of charges, +Q and –Q (as electric dipole), where the dashed line shows the location of zero potential surface. If we slide a conductive plane over the zero potential surface, the field lines in the upper half plane are unchanged.

Note that the charge can be in any distribution (point charge, line charge, surface or volume charge) and the image charge is a mirror image of opposite polarity.A monopole antenna is excited by a current source at its base. By image theory, the current in the image will be the same with the current in actual monopole. The pair of monopole resembles a dipole antenna.A monopole antenna placed over a conductive plane and half the length of a corresponding dipole antenna will have identical field patterns in the upper half plane.


For the upper half plane (00< θ<900), the time averaged power, max power density and normalized power density for the quarter wave monopole is the same with half wave dipole. But the pattern solid angle is different.


Since the normalized power density is zero for (900< θ<1800), the pattern solid angle:

dPnp

Integrate over all space will be half value of Ωp for half wave dipole. So, for quarter wave monopole,

28.34829.3 max

p

p D

See that the radiation resistance is halved, and the antenna impedance :

6.3630

pradR

25.216.36 jZant


Summary of Key Antenna Parameters

4.5 Antenna ArraysThe antennas we have studied so far have all been omnidirectional – no variation in φ. A properly spaced collection of antennas, can have significant variation in φ leading to dramatic improvements in directivity.

A Ka-Band Array Antenna

Antenna Arrays (Cont’d..)An antenna array can be designed to give a particular shape of radiating pattern. Control of the phase and current driving each array element along with spacing of array elements can provide beam steering capability. For simplification: All antenna elements are identical The current amplitude is the same feeding each

element. The radiation pattern lies only in xy plane, θ=π/2 The radiation pattern then can be controlled by: controlling the spacing between elements or controlling the phase of current driving for each element

Antenna Arrays (Cont’d..)For simple example, consider a pair of dipole antennas driven in phase current source and separated by λ/2 on the x axis. Assume each antenna radiates

independently, at far field point P, the fields from 2 antennas will be 180 out-of-phase, owing to extra λ/2 distance travel by the wave from the farthest antenna fields cancel in this direction. At point Q, the fields in phase and adds. The E field is then twice from single dipole, fourfold increase in power broadside array max radiation is directed broadside to axis of elements.

Antenna Arrays (Cont’d..)

Modify with driving the pair of dipoles with current sources 180 out of phase. Then along x axis will be in phase and along y axis will be out of phase, as shown by the resulting beam pattern endfire array max radiation is directed at the ends of axis containing array elements.


Pair of Hertzian Dipoles

Recall that the far field value of E field from Hertzian dipole at origin,

aE sin

400 RkeI

jjkR

sS

But confining our discussion to the xy plane where θ = π/2,

aE

RkeI

jjkR

sS 400


Consider a pair of z oriented Hertzian dipole, with distance d, where the total field is the vector sum of the fields for both dipoles and the magnitude of currents the same but a phase shift between them.

aaEEE

2

20

1

1020100 44

21

RkeI

jR

keIj

jkRs

jkRs

SStotS

Where, jss eIIII 0201

Antenna Arrays (Cont’d..)Assumption,

2121 RrR

And,

Antenna Arrays (Cont’d..)Where geometrically we could get,

cos2

cos2 21

drRdrR

Thus, the total E field becomes:

aE

2cos

22cos

22000 4

dkjdkjj

jkR

totS eeeR

keIj

With Euler’s identity, the total E field at far field observation point from two element Hertzian dipole array becomes :

aE

2cos

2cos2

420

00dke

RkeI

jjjkR

totS

Antenna Arrays (Cont’d..)To find radiated power,

r

rtotSSS

dkR

kI

Er

a

aHEP

2cos

2cos4

32

21Re

21,

2,

222

22200

200

*

It can be written as:

rarrayunit FFr a P

,

2,


Unit factor, Funit is the max time averaged power density for an individual antenna element at θ=π/2

22

22200

32 R

kIFunit

An array factor, Farray is

2cos4 2

arrayF

Where, coskd

This is the pattern function resulting from an array of two isotropic radiators.

94

Example 5

The λ/2 long antennas are driven in phase and are λ/2 apart. Find: the far field radiation

pattern for a pair of half wave dipole shown.

the maximum power density 1 km away from the array if each antenna is driven by a 1mA amplitude current source at 100 MHz.

Solution to Example 5At 100 MHz, λ = 3m, so that 1 km away is definitely in far field. For a half wave dipole, we have :

rrr

IrPr a aP

2

2

2

20

sin

cos2

cos15

,,

The unit factor can be found by evaluating above at θ = π/2 ,

2

2015

r

IFunit

Solution to Example 5 (Cont’d..)The array factor, with d = λ/2 and =0 (due to antennas are driven at the same phase) :

cos

2cos4 2

arrayF

For the array, we now have :

cos

2cos

60,

2, 2

2

20

r

IFFr rarrayunit a P

The normalized power function is:

cos2

cos,

2,

,2

,,

22

maxrP

rP

nP

Solution to Example 5 (Cont’d..)This can be plotted as :

But how to plot?!?!?!

The maximum radiated power density at 1000m is :

22

23

2

20

max 191000

106060

m

pW

r

IP

Use MATLAB!!


N - Element Linear Arrays

The procedure of two-element array can be extended for an arbitrary number of array elements, by simplifying assumptions : The array is linear antenna elements are evenly spaced,

d along a line. The array is uniform each antenna element driven by

same magnitude current source, constant phase difference, between adjacent elements.


10

2030201 ,....,, Nj

sNj

sj

ss eIIeIIeIIII


The far field electric field intensity :

rNjjj

jkR

totS eeeR

keIj aE

120

00 ...14

Where, coskd

Manipulate this series to get:

2sin

2sin

2

2 N

FarrayWith the max value as :

2max

NFarray


So then the normalized power pattern for these elements is :

2sin

2sin

12

2

2max

N

NF

FP

array

arrayn

102

Example 6

Five antenna elements spaced λ/4 apart with progressive phase steps 300. The antennas are assumed to be linear array of z oriented dipoles on the x axis. Find: the normalized radiation pattern in xy plane the plot of the radiation pattern.


To find the array factor, first need to find psi, Ψ:

6cos

218030cos

42

cos

kd

Inserting this ratio to array factor,

12cos

4sin

125cos

45sin

2sin

2sin

2

2

2

2

N

Farray 252max

NFarray

Solution to Example 6 (Cont’d..)The normalized radiation pattern is :

12cos

4sin

125cos

45sin

251

2

2

nP

But how to plot?!?!?!

Use MATLAB!!

The plot is :


Parasitic Arrays

Not all the elements in array need be directly driven by current source. A parasitic array typically one driven element and several parasitic elements, the best known parasitic array is Yagi-Uda antenna.


On one side of the driven element is reflector length and spacing are chosen to cancel most of the radiation in that direction, as well as to enhance the direction to forward or main beam direction.

Several directors (four to six) focus the main beam in the forward direction high gain and easy to construct.

Parasitic elements tend to pull down the Rrad of the driven element. E.g. Rrad of dipole would drop from 73Ω to 20Ω when used as the driven element in Yagi-Uda antenna.But higher Rrad is more efficient, so use half wavelength folded dipole antenna (four times Rrad of half-wave dipole!)

4.6 Helical Antennas

Diameter of ground plane at least 3λ/4

Helical Antennas (Cont’d..)

Helical Antennas (Cont’d..)Important Parameters


For this special case,

The radiated field is circularly polarized in all directions other than θ = 00


Parameters for End Fire mode

Helical Antennas (Cont’d..)Feed Design for Helical Antennas

4.7 Yagi Uda Array Antennas

Yagi Uda Array Antennas (Cont’d..)

124

Example 7

With above parameters, design a Yagi Uda Array antenna by finding the element spacing, lengths and total array length.

Given


from given directivity = 9.2 dB

Solution to Example 7(Cont’d..)


Therefore,Total array length: 0.8 λThe spacing between directors: 0.2 λThe reflector spacing: 0.2 λThe actual elements length:

L3 = L5 : 0.447λ

L4 : 0.443λ

L1 : 0.490λ

128


129

Fig 10.25

Solution to Example 7(Cont’d..)Design Curves to Determine Element Lengths

X

.

130

Fig 10.26


Yagi Uda Array Antennas (Cont’d..)

4.8 Antennas for Wireless CommunicationsParabolic Reflectors

Parabolic dish antenna

Parabolic reflector antenna

Antennas for Wireless Communications (Cont’d..)

Cassegrain reflector antenna

All of these parabolic reflectors operates based on the geometric optics principle that a point source of radiation placed at the focal point of parabolic reflector will radiate the energy incident on the dish in a narrow and collimated beam.For high efficient, the dish must be significantly larger

than the radiation wavelength, and has a directive feed.

Antennas for Wireless Communications (Cont’d..)Patch Antennas

Other shapes such as circles, triangles and annular rings also been used. It can be excited by an edge or probe fed, where its location is chosen for impedance match between cable and antenna.

Antennas for Wireless Communications (Cont’d..)Folded Dipole Antennas

A pair of half-wavelength dipole elements are joined at the ends and fed from the center of one of the pair. If the two sections are close together (d on the order of λ/64), the impedance will be four times greater than the regular λ/2 dipole antenna. The directivity is the same but the bandwidth is significantly broader.

AntennaEnd

chptr 4 antenna dipole yagi2015

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